Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $y = \dfrac{x^2 - 49}{-x^2 + 49} \times \dfrac{x - 6}{-x + 7} $
Solution: First factor out any common factors. $y = \dfrac{x^2 - 49}{-(x^2 - 49)} \times \dfrac{x - 6}{-(x - 7)} $ Then factor the quadratic expressions. $y = \dfrac {(x + 7)(x - 7)} {-(x + 7)(x - 7)} \times \dfrac {x - 6} {-(x - 7)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac { (x + 7)(x - 7) \times (x - 6)} { -(x + 7)(x - 7) \times -(x - 7)} $ $y = \dfrac {(x + 7)(x - 7)(x - 6)} {(x + 7)(x - 7)(x - 7)} $ Notice that $(x + 7)$ and $(x - 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {\cancel{(x + 7)}(x - 7)(x - 6)} {\cancel{(x + 7)}(x - 7)(x - 7)} $ We are dividing by $x + 7$ , so $x + 7 \neq 0$ Therefore, $x \neq -7$ $y = \dfrac {\cancel{(x + 7)}\cancel{(x - 7)}(x - 6)} {\cancel{(x + 7)}(x - 7)\cancel{(x - 7)}} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $y = \dfrac {x - 6} {x - 7} $ $ y = \dfrac{x - 6}{x - 7}; x \neq -7; x \neq 7 $